11) lg_4 lg_4 lg_2 16−lg_2 lg_2 (√3) 12. (5lg_3 3−lg_4 1)^2 +(((1/(lg_2 8))×lg_3 27)/(lg


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Question Number 58896 by cesar.marval.larez@gmail.com last updated on 03/May/19

$11) \lg_{4} \lg_{4} \lg_{2} 16 - \lg_{2} \lg_{2} \sqrt{3}$ $12 . {({5 \lg}_{3} 3 - \lg_{4} 1)}^{2} + \frac{\frac{1}{\lg_{2} 8} \times \lg_{3} 27}{\lg_{\sqrt{2}} \frac{1}{2}}$
	Answered by tanmay last updated on 01/May/19

	${l g}_{4} {l g}_{4} {l g}_{2} 16$ ${l g}_{4} {l g}_{4} {l g}_{2} 2^{4}$ ${l g}_{4} {l g}_{4} 4 [{l g}_{2} 2 = 1]$ ${l g}_{4} 1$ $= 0$ ${l g}_{2} {l g}_{2} 3^{\frac{1}{2}}$ ${l g}_{2} 2^{- 1} {l g}_{2} 3$ $= (- 1) \times {l g}_{2} 3$ $a n s = 0 - (- {l g}_{2} 3) = {l g}_{2} 3$
	Answered by tanmay last updated on 01/May/19

	${(5 {l g}_{3} 3 - {l g}_{4} 1)}^{2}$ $= {(5 \times 1 - 0)}^{2} = 25$ $\frac{\frac{1}{{l g}_{2} 8} \times {l g}_{3} 27}{{l g}_{\sqrt{2}} \frac{1}{2}}$ $= \frac{\frac{1}{{l g}_{2} 2^{3}} \times {l g}_{3} 3^{3}}{{l g}_{\sqrt{2}} {(\sqrt{2})}^{- 2}}$ $= \frac{\frac{3}{3}}{- 2} = \frac{- 1}{2}$ $a n s i s 25 - \frac{1}{2}$ $= \frac{49}{2}$
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